A train leaves a station and travels north at a speed of 30 km/h. Five hours later, a second train leaves on a parallel track and travels north at 105 km/h. How far from the station will they meet?

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  210 km

Step-by-step explanation:

The position of the first train as a function of the time after it left is ...

  f(t) = 30t

The position of the second train as a function of time after the first train left is ...

  s(t) = 105(t -5)

The positions are equal when ...

  30t = 105(t -5)

  525 = 75t . . . . . . add 525-30t to both sides

  7 = t . . . . . . . . . divide by 75

  f(7) = s(7) = 30(7) = 210 = 105(7 -5)

The trains will meet 210 km from the station.

Alternate solution

The first train has a 5-hour head start, amounting to 5×30 = 150 km. The second train catches up at the rate of (105 -30) = 75 km/h. So, the second train covers the 150 km gap in 150/75 = 2 hours. It will have traveled 105×2 = 210 km in that time. (Effectively, we're using the start time of the second train as the time reference. 30(t+5) = 105t)

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

7.3

step-by-step explanation: